Surface of a sphere I have this picture, and I have to explain why the lower part of the sphere corresponds to the representative surface of a function. I don't manage to understand the goal of the question, anybody can help me?
A second question is to find the previous function. I manage to find $$f(x,y) = R - \sqrt{R^2 -x^2 - y^2}$$
But I don't find how to show that $$\iint_\mathscr D f(x,y)dxdy=\frac{πR^3}{3}$$ 
Somebody has starter idea ?
 A: The image is a bit unclear, as it doesn't give you any real reference. But I just assume (as I see) that it is a sphere which south pole is on the origin. In which, case your function is correct.  
To answer the first question, you should ask yourself: when a surface represents a function? When to a point $(x,y)$ corresponds at most one $z$. Graphically, you may represent this by saying that vertical lines intersect the surface no more than once.
(If you ever did something similar for functions of one variable, you remember that you had to intersect the graph at most once with vertical lines; it's the same idea but in three dimensions)
For the second question, you may solve it in two ways:


*

*Using elementary geometry: you just have to calculate the difference of the volumes of the cylinder, with the same radius as the sphere and height which goes from the origin of the axis to the center of the sphere (i.e. equals to the radius of the sphere), and the lower hemisphere. This yields $\pi R^3 - \frac{2}{3}\pi R^3 = \frac{\pi}{3} R^3$

*Using calculus: I won't do all calculations for this but it shouldn't be too hard. Just use the polar coordinates and it should do the trick. Just remember to find the right boundaries for $\rho$ which will depend on the height $z$ of course. 

A: I guess, the second question is meant to be a straightforward calculation of the integral
$$
\iint_{x^2+y^2\le R^2}\left(R-\sqrt{R^2-x^2-y^2}\right)\,dxdy.
$$ 
(It has nothing to do with the surface of the sphere, but is the volume between the hemisphere and the $xy$-plane.) In the polar coordinates
$$
\begin{cases}
x=r\cos\phi,\\
y=r\sin\phi
\end{cases}
$$
the domain of integration $x^2+y^2\le R^2$ is described as $0\le r\le R$, $0\le\phi\le 2\pi$. Change the variables (and do not forget the Jacobian $=\color{red}r$)
$$
\int_0^R\int_0^{2\pi}\left(R-\sqrt{R^2-r^2}\right)\color{red}r\,d\phi dr=2\pi\int_0^R\left(R-\sqrt{R^2-r^2}\right)\,\frac{dr^2}{2}=\pi\int_0^{R^2}\left(R-\sqrt{R^2-t}\right)\,dt=...=\frac{\pi R^3}{3}.
$$
